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Given the action (note $G_{ab}$ is a symmetric matrix, i.e. $G_{ba} = G_{ab}$):

$$S = \int dt \Big( \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)\Big)$$

Show (using Euler Lagrange's equation) that the following equation holds:

$$\ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c = -\sum_{a} F^{da}\partial_a V$$

Where $F^{ab}$ is the inverse of $G_{ab}$.

Also note that:

$$\sum_{b} F^{ab}G_{ab} = \delta_c^a$$

$$\partial_{a} = \frac{\partial}{\partial q_a}$$

What I have done:

We know that the action functional corresponds to the Lagrangian (for the time interval $[t_0, t_1]$):

$$S[q] = \int_{t_0}^{t_1} L(q, \dot q, t)dt$$

Thus:

$$L = \sum_{ab} G_{ab} \dot q^a\dot q^b-V(q)$$

Euler Lagrange's equation is:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{\partial L}{\partial q_k}$$

Let's go step by step:

1) We compute the term $\frac{\partial L}{\partial \dot q_k}$ (which turns out to be the definition of generalized momentum):

$$p_k = \frac{\partial L}{\partial \dot q_k} = \sum_{ab} \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = \sum_{b} G_{kb} \dot q^b + \sum_{a} G_{ak} \dot q^a = 2\sum_a G_{ak} \dot q^a$$

2) We now compute the term $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big)$

NOTE: I know that the symmetric matrix $G_{ab}$ only depends on $q_k$. By the chain rule (for the sake of clarity: $G_{ab} (q_k)$ notation means that the matrix $G_{ab}$ is a function of $q_k$):

$$\frac{d}{dt} \sum_a G_{ab} (q_k) = \sum_{a} \partial_k G_{ab} \dot q_k$$

That being said, let's go through the calculation:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q_k }\Big) = \frac{d}{dt}\Big( 2\sum_a G_{ak} (q_k) \dot q^a \Big) = 2\sum_a G_{ak} \ddot q^a + 2\sum_a \partial_k G_{ak} \dot q^a \dot q^k$$

3) We now compute the term $\frac{\partial L}{\partial q_k}$

$$\frac{\partial L}{\partial q_k} = \sum_{ab} \partial_k G_{ab} \dot q^a\dot q^b - \partial_k V(q)$$

Here's where I get stuck: I do not see why the following holds:

$$2\sum_a F^{da}G_{ak} \ddot q^a + 2\sum_a F^{da}\partial_k G_{ak} \dot q^a \dot q^k -\sum_{ab} F^{da}\partial_k G_{ab} \dot q^a\dot q^b + F^{da}\partial_k V(q) = \ddot q^d + \frac{1}{2}\sum_{abc}F^{da}\Big(\partial_cG_{ab} + \partial_bG_{ac} - \partial_a G_{bc}\Big)\dot q^b\dot q^c + \sum_{a} F^{da}\partial_a V$$

Please let me know if need more details in the above equation; I can provide more steps.

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  • $\begingroup$ Does $G_{ab}$ have an explicit expression? $\endgroup$ Dec 9, 2019 at 16:29
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    $\begingroup$ This might be helfpul: physics.stackexchange.com/q/137422 . To simplify things, also check out en.wikipedia.org/wiki/Christoffel_symbols, specifically those of the second kind. $\endgroup$
    – R. Romero
    Dec 9, 2019 at 16:45
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    $\begingroup$ Modify Newton's First law of motion to be : Objects travel along geodesics in space-time in the absence of a net force, then derive the geodesic equations. The techniques used there can be used to re derive laws of motion in the presence of a force my modifying the Lagrangian appropriately: people.uncw.edu/hermanr/GR/geodesic.pdf $\endgroup$
    – R. Romero
    Dec 9, 2019 at 16:49
  • $\begingroup$ @Tesseract all I know about $G_{ab}$ is that it is symmetric and depends on $q_{k}$ $\endgroup$
    – JD_PM
    Dec 9, 2019 at 17:05

1 Answer 1

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A hint: when you use the chain rule, you need a new index, so $\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial q^k}\right)$ would be $$2\sum_aG_{ak}\ddot{q}^a+2\sum_{ac}\partial_cG_{ak}\dot{q}^a\dot{q}^c$$ We can split this up into parts, and relabel indices to get $$2\sum_aG_{ak}\ddot{q}^a+\sum_{ac}\partial_cG_{ak}\dot{q}^a\dot{q}^c+\sum_{ac}\partial_aG_{ck}\dot{q}^a\dot{q}^c$$ Can you continue from here? Also, check out this article on Einstein summation notation; it will make your work cleaner and more readable!

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